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4 times hexagonal numbers: a(n) = 4*n*(2*n-1).
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%I #64 Oct 19 2022 15:55:21

%S 0,4,24,60,112,180,264,364,480,612,760,924,1104,1300,1512,1740,1984,

%T 2244,2520,2812,3120,3444,3784,4140,4512,4900,5304,5724,6160,6612,

%U 7080,7564,8064,8580,9112,9660,10224,10804,11400,12012,12640,13284

%N 4 times hexagonal numbers: a(n) = 4*n*(2*n-1).

%C a(n) also can represented as n concentric squares (see example). - _Omar E. Pol_, Aug 21 2011

%C Sequence found by reading the line from 0, in the direction 0, 4, ..., in the square spiral whose vertices are the triangular numbers A000217. - _Omar E. Pol_, Sep 03 2011

%H Ivan Panchenko, <a href="/A085250/b085250.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = A067239(n)/2, for n>0.

%F Sum_{n>0} 1/a(n) = log(2)/2.

%F a(n) = A000384(n)*4. - _Omar E. Pol_, Dec 11 2008

%F a(n) = 16*n + a(n-1) - 12 (with a(0)=0). - _Vincenzo Librandi_, Aug 08 2010

%F G.f.: 4*x*(1 + 3*x)/(1 - 3*x + 3*x^2 - x^3). - _Colin Barker_, Jan 04 2012

%F E.g.f.: 4*x*(2*x + 1)*exp(x). - _G. C. Greubel_, Jul 14 2017

%F a(n) = A046092(2n-1), for n > 0. - _Bruce J. Nicholson_, Sep 04 2017

%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 - log(2)/4. - _Amiram Eldar_, Mar 17 2022

%e From _Omar E. Pol_, Aug 21 2011: (Start)

%e Illustration of initial terms as concentric squares:

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%e . 4 24 60

%e .

%e (End)

%t Table[8*n^2 - 4*n, {n, 0, 50}] (* _G. C. Greubel_, Jul 14 2017 *)

%t 4 PolygonalNumber[6,Range[0,50]] (* _Harvey P. Dale_, Oct 19 2022 *)

%o (PARI) a(n)=4*n*(2*n-1) \\ _Charles R Greathouse IV_, Sep 24 2015

%Y Cf. A067239, A000384, A033581, A046092, A152734, A152751, A194274.

%K nonn,easy

%O 0,2

%A _Gary W. Adamson_, Jun 23 2003

%E Edited by _Don Reble_, Nov 13 2005

%E Added zero, better definition, corrected offset and edited original formula. - _Omar E. Pol_, Dec 11 2008